5% solution of cane sugar in water has a freezing point of 271 kelvin. cal the freezing point of 5% glucose in water. the freezing point of water is taken as 273.15k

Asked by sg | 7th Apr, 2013, 12:50: PM

Expert Answer:

Here, ΔTf = (273.15 − 271) K

= 2.15 K

Molar mass of sugar (C12H22O11) = 12 × 12 + 22 × 1 + 11 × 16

= 342 g mol−1

5% solution (by mass) of cane sugar in water means 5 g of cane sugar is present in (100 − 5)g = 95 g of water.

Now, number of moles of cane sugar

= 0.0146 mol

Therefore, molality of the solution,

= 0.1537 mol kg−1

Applying the relation,

ΔTf = Kf × m

= 13.99 K kg mol−1

Molar of glucose (C6H12O6) = 6 × 12 + 12 × 1 + 6 × 16

= 180 g mol−1

5% glucose in water means 5 g of glucose is present in (100 − 5) g = 95 g of water.

Number of moles of glucose

= 0.0278 mol

Therefore, molality of the solution,

= 0.2926 mol kg−1

Applying the relation,

ΔTf = Kf × m

= 13.99 K kg mol−1 × 0.2926 mol kg−1

= 4.09 K (approximately)

Hence, the freezing point of 5% glucose solution is (273.15 − 4.09) K= 269.06 K.



Answered by  | 10th Apr, 2013, 09:56: AM

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